Carbon disulfide is prepared by reacting carbon w/ sulfur dioxide according to 5C + 2SO2 → CS2 + 4CO. 7.5 g of carbon react with 7.5 g if Sulfur dioxide, how many gram of CS2 are produced

Answer 1

5C + 2SO2 → CS2 + 4CO

mass C = 7.5g mass SO2 = 7.5g mass CS2 produced =??

When given masses of both reactants, we must check for a limiting reactant. We can only calculate the products made by using the mass of the limiting reactant. Luckily, the process of finding the limiting reactant leads us toward the mass of the products anyway so it’s not much extra work.

Process:

convert grams reactants —> moles reactants

convert moles reactants —> moles CS2 produced

choose limiting reactant

convert moles CS2 made by limiting reactant —> grams CS2

Convert grams reactants —> moles reactants

For this step we need to calculate the molar mass using a periodic table. Find the molecular weights of C, S, and O:

C: 12.011 g/mol

S: 32.066 g/mol

O: 15.998 g/mol

(Remember this will not involve the coefficients of the balanced equations, only any subscripts!)

Molar mass of reactant 1:

C * 1 = 12.011 g/mol

Molar mass reactant 2:

S * 1 + O * 2 = 64.062 g/mol

Convert g—>mol

grams / molar mass = moles

7.5g C / 12.011 g/mol C = 0.62 mol C

7.5 g SO2 / 64.062 g/mol SO2 = 0.12 mol SO2

Next we use the mole ratio (the coefficients of the balanced equation) to convert to moles of product. We want to find how much CS2 is produced by each reactant separately.

mol reactant * coeff product/coeff reactant = moles product

0.62 mol C * 1CS2 / 5C = 0.12 mol CS2 produced by C

0.12 mol SO2 * 1CS2 / 2SO2 = 0.060 mol CS2 produced by SO2

Since SO2 made the least amount of product, it is the limiting reactant. We won’t produce 0.12 mol CS2 because SO2 limits us to making only 0.060 mol CS2. We can only use 0.06 mol for future calculations.

Lastly, convert moles to grams or product.

moles * molar mass = grams

molar mass of CS2:

C*1 + S*2 = 12.011 + 64.132 = 76.143 g/mol

0.060 mol CS2 * 76.143 g/mol = 4.6 grams CS2 produced